If $Y|X \sim N(0,1)$, does that mean $Y$ itself is also $N(0,1)$?

It looks like it as I can write $$ E[Y] = E[E[Y|X]] = E[0] = 0, $$ and $$ Var[Y] = Var[E[Y|X]] + E[Var[Y|X] = Var[0] + E[1] = 1. $$ But although I get the correct expectation and variance I don't see how we can say $Y$ is normally distributed?

If it is true, does this only hold because we are dealing with the normal distribution?

  • 2
    $\begingroup$ $Y|X$ has a distribution that is (a) independent from $X$ and (b) a Normal distribution. The conclusion does not involve any specifics of the Normal distribution. $\endgroup$
    – Xi'an
    Commented Nov 26, 2020 at 18:20
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    $\begingroup$ Expression for marginal of $Y$: $Pr(Y \leq y) = \int_{\mathbb R} \Phi(y) f(x) dx$. Regardless of what $f(x)$ is it will integrate to one to give that $Pr(Y \leq y) = \Phi(y)$. So it's normal. $\endgroup$
    – Dayne
    Commented Nov 26, 2020 at 18:27
  • $\begingroup$ @Dayne What is $\Phi(y)$? And where did you get that expression for the probability of $Y$ being less than $y$. The definition of $Pr(Y \le y)$ is $Pr(Y \le y) = \int_{-\infty}^y f_Y(t) dt$ where $f_Y$ is the probability density function of $Y$. $\endgroup$
    – csss
    Commented Nov 26, 2020 at 19:58
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    $\begingroup$ $\Phi(y)$ is the cdf for standard normal. Since, $Y|X$ is standard normal, the marginal can be calculated as shown. $\endgroup$
    – Dayne
    Commented Nov 26, 2020 at 20:06